Determine the equivalent resistance for parallel resistors.

The resistors $R$ and $X$ are connected in parallel. The equivalent resistance $R_{eq}$ is given by:

$$R_{eq} = \frac{R \cdot X}{R + X}$$

The total voltage $E$ across the parallel combination is constant. According to Ohm's Law, $E = I \cdot R_{eq}$, where $I$ is the total current measured by the ammeter.
Substituting the expression for $R_{eq}$:

$$E = I \cdot \frac{R \cdot X}{R + X}$$

When $R = 5 \, \Omega$, the current $I = 3 \, A$.

$$E = 3 \, A \cdot \frac{5 \, \Omega \cdot X}{5 \, \Omega + X}$$ $$E = \frac{15X}{5 + X} \quad (*)$$

When $R = 2 \, \Omega$, the current $I = 6 \, A$.

$$E = 6 \, A \cdot \frac{2 \, \Omega \cdot X}{2 \, \Omega + X}$$ $$E = \frac{12X}{2 + X} \quad (**)$$

Since $E$ is constant, we can set equation $(*)$ equal to equation $(**)$:

$$\frac{15X}{5 + X} = \frac{12X}{2 + X}$$

Assuming $X \neq 0$ (as $X$ is a resistance), we can divide both sides by $X$:

$$\frac{15}{5 + X} = \frac{12}{2 + X}$$

Cross-multiply:

$$15(2 + X) = 12(5 + X)$$ $$30 + 15X = 60 + 12X$$

Subtract $12X$ from both sides:

$$30 + 15X - 12X = 60$$ $$30 + 3X = 60$$

Subtract 30 from both sides:

$$3X = 60 - 30$$ $$3X = 30$$

Divide by 3:

$$X = \frac{30}{3}$$ $$X = 10 \, \Omega$$

The value of $X$ is $10 \, \Omega$. This corresponds to option C.

The final answer is $\text{C. 10 ohms}$.